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A LOCAL GROTHENDIECK DUALITY THEOREM FOR COHEN-MACAULAY IDEALS JOHANNES LUNDQVIST Abstract. WegiveanewproofofarecentresultduetoMatsAn- 2 derssonandElizabethWulcan,generalizingthelocalGrothendieck 1 duality theorem. It canalsobe seen asa generalizationof a previ- 0 ousresultbyMikaelPassare. Ourmethoddoesnotrequiretheuse 2 of the Hironaka desingularization theorem and it provides a semi- n explicit realization of the residue that is annihilated by functions a from the given ideal. J 3 2 ] V 1. Introduction C . Let O be the ring of germs of holomorphic functions at 0 ∈ Cn h 0 at and let Ωn0 denote the germs of holomorphic (n,0)-forms. The ring O0 m is Noetherian and hence all ideals J ⊂ O0 will be finitely generated. Assume first that J is generated by n functions f = (f ,...,f ) and [ 1 n that their common zero set consists of one single point, the origin. 1 v Then the Grothendieck residue, Resf, is defined as 9 n 4 1 ξ(z) 47 (1) Resf(ξ) = (cid:18)2πi(cid:19) Z|fi(z)|=ǫ f1(z)···fn(z), ξ ∈ Ωn0, . 1 and is independent of ǫ. Observe that we can multiply Res with a 0 f 2 holomorphic germ ϕ by letting ϕRes (ξ) = Res (ϕξ). There is a well f f 1 known theorem, see for example [9], saying that J is equal to the : v annihilator ideal of Res , i.e., f i X (2) ϕRes (ξ) = 0, ∀ξ ∈ Ωn, iff ϕ ∈ J. r f 0 a We will refer to that theorem as the local Grothendieck duality theorem. There is a cohomological interpretation of the Grothendieck residue. Let Ω be an open neighborhood of 0 such that f , j = 1,...,n, and j ξ are defined there. Let D = {z;f (z) = 0} and U = Ω\ D . Then j j j j ξ/f ...f can be considered as an (n−1)-cochain for the sheaf of holo- 1 n morphic(n,0)-formsandthecovering{U } ofΩ\{0}. Sincethere j j=1,...,n areno(n−1)-coboundaries, ξ/f ...f defines aCˇech cohomologyclass 1 n andbytheDolbeaulttheorem, [9], wegetaDolbeaultcohomologyclass ωξ of bidegree (n,n−1). The Grothendieck residue can now be rewrit- ten as integration of ωξ over the boundary of a small neighborhood, D, 1 2 JOHANNESLUNDQVIST of the origin, (3) Res (ξ) = ωξ. f Z ∂D A proof of this can be seen in [9] where one can also see a proof of the fact that the class ω can be represented by the explicit form 1 n (−1)i−1f¯df¯ ∧...∧df¯ ∧...∧df¯ ∧ξ ωξ = n! i 1 i n , (cid:18)2πi(cid:19) P (|f |2 +...+|f |2)n 1 cn ¯ ¯ where df means that df is omitted. i i Assume now that the ideal J is generated by p functions f ,...,f c 1 p and that we do not have any restrictions on the common zero set Z. With the use of Hironaka’s desingularization theorem one can define a residue current 1 1 ξ(z) ¯ ¯ (4) ∂ ∧...∧∂ ·ξ = lim , fp f1 δ→0Z|fi(z)|=ǫi(δ) f1(z)···fp(z) for smooth test forms ξ. In order for the limit to exist it has to be taken over a so called admissible path meaning that ǫ (δ) tends faster i to zero than any power of ǫ (δ). The current (4) is called the Coleff- i+1 Herrera product and was defined in [5]. In the special case of p = n and Z = {0} we get the Grothendieck residue if we restrict the Coleff- Herrera product to the holomorphic germs. In [7] and [10] Dickenstein-Sessa and Passare independently proved that the Coleff-Herrera product satisfies the duality theorem, i.e, that the annihilator ideal of (4) is equal to J, in the case when J defines a complete intersection. That is, the case when the codimension of J is equal to p. Passare also defines a cohomological residue satisfying the duality theorem in that case generalizing the Grothendieck duality theorem to complete intersections. In [2] Andersson and Wulcan con- struct a residue current that satisfies a duality theorem for arbitrary ideals and coincides with the Coleff-Herrera product if the ideal defines a complete intersection. They also show that the residue can be expressed cohomologically in the Cohen-Macaulay case. In this paper we find a new proof of the result of Andersson and Wulcan in the Cohen-Macaulay case avoiding using Hironaka desingularization used in [2]. The residue is similar to (3) and is obtained from a double complex defined from a free resolution of J. In the special case of a complete intersection it coincides with the cohomological residue in [10]. 2. Set up and statement Remember that a local Noetherian ring R is called Cohen-Macaulay if the maximal length of a regular sequence in R is equal to the di- mension of R. An ideal J ⊂ R is called Cohen-Macaulay if R/J is 3 Cohen-Macaulay. All ideals inO whose variety iszero-dimensional are 0 Cohen-Macaulay. Also, all ideals in O that define a complete inter- 0 section are Cohen-Macaulay but the converse is not true. For example, the ideal hz2,zw,w2i ⊂ O is Cohen-Macaulay (because its variety is 0 zero-dimensional) but do not define a complete intersection. Assume that the common zero set Z of f ,...,f ∈ O has pure 1 m 0 codimension p and that J = hf ,...,f i is Cohen-Macaulay. The fact 1 m that J is Cohen-Macaulay is equivalent (because of the Auslander- Buchsbaum formula [8]) to the existence of a minimal free resolution of O /J 0 (5) 0 → O⊕rp →fp O⊕rp−1 f→p−1 ... →f2 O⊕r1 →f1 O → O /J → 0 0 0 0 0 0 having length p. Here f1 can be represented as the row-matrix where the the i:th column is f and fk,k > 1, are matrices with holomorphic i functions as entries. Oka’s lemma, [9], implies that there exists a small neighborhood Ω around 0 such that the complex (6) 0 → O⊕rp →fp O⊕rp−1 f→p−1 ... →f2 O⊕r1 →f1 O → O /J → 0 z z z z z is exact for all z ∈ Ω. If we let E be a trivial vector bundle of rank r over Ω we get an j j induced complex of trivial vector bundles fp fp−1 f2 f1 (7) 0 → E → E → ... → E → E → 0. p p−1 1 0 Note that O /J = 0 if z ∈ Ω\Z and that the complex (7) is pointwise z exact there. Indeed, assume that k < p and that (z ,x) ∈ (Ω\Z)×Crk 0 is a point such that fk(z )x = 0. Note first that there exist a non-zero 0 function ϕ ∈ O⊕rk such that fkϕ = 0 because otherwise Kerfk = {0} z0 and hence k = p which is a contradiction since we assumed that k < p. Take such a ϕ. We know from the exactness of (6) that there exists ψ ∈ O⊕rk+1 such that fk+1ψ = ϕ. By scaling we can assume that z0 ϕ(z ) = x and by choosing y = ψ(z ) we get that the point (z ,y) ∈ 0 0 0 (Ω\Z)×Crk+1 is mapped to (z ,x). 0 The exactness of (7) and a simple induction over k shows that fk has constant rank in Ω \ Z and thus Kerfk is a sub-bundle of E . k Since fk+1 is a pointwise surjection to the sub-bundle Kerfk we get that the corresponding complex of smooth sections is exact. We have just proved the following proposition. Proposition 2.1. Let E (Ω,E ) denote the set of smooth 0,q k (0,q)-sections of E over Ω. With fk,E ,Ω and Z as above, the com- k k plex fp fp−1 0 → E (Ω\Z,E ) → E (Ω\Z,E ) → ... 0,q p 0,q p−1 f1 ... → E (Ω\Z,E ) → 0 0,q 0 is exact for all q. 4 JOHANNESLUNDQVIST We are now ready to define the complex that will give us the cohomology classes we need in order to state the main theorem. The operators fj and ∂¯ define the double complex (8) . . . . . . ∂¯ ∂¯ x x fk+1  fk  fk−1 ... −−−→ E (Ω\Z,E ) −−−→ E (Ω\Z,E ) −−−→ ... 0,q+1 k 0,q+1 k−1 . ∂¯ ∂¯ x x fk+1  fk  fk−1 ... −−−→ E (Ω\Z,E ) −−−→ E (Ω\Z,E ) −−−→ ... 0,q k 0,q k−1 ∂¯ ∂¯ x x . . . . . . Let L be the total complex of (8), i.e., (9) L : ... ∇→f Lr(Ω\Z) →∇f Lr+1(Ω\Z) →∇f .... where Lr(Ω\Z) = E (Ω\Z,E ) 0,k+r k Mk and ∇ : Lr(Ω\Z) → Lr+1(Ω\Z) is defined as ∇ = f −∂¯. f f Here f should be interpreted as (−1)qfk on E (Ω\Z,E ). We know 0,q k fromProposition2.1thatthedoublecomplex(8)hasexactrowsandby a standard spectral sequence argument we see that the total complex L is exact. Now, let ϕ ∈ O . Then we can view ϕ as an element in L0(Ω\ Z) 0 for some Ω such that the complex L is exact. Moreover, ∇ ϕ = 0 so f there exists an element v in L−1(Ω\Z) such that ∇ v = ϕ. If we write f v = v +...+v where v ∈ E (Ω\Z,E ) we see that fv = ϕ and 1 p k 0,k−1 k 1 ¯ ¯ fv = ∂v for j = 2,...p. Especially we get that ∂v = 0. Now, if j j−1 p v,w ∈ L−1(Ω\Z) are such that ∇ v = ∇ w = ϕ, then there exists an f f element u ∈ L−2(Ω\Z) such that ∇ u = v−w and thus ∂¯u = v −w . f p p p ¯ This means that v (a vector of r ∂-closed smooth (0,p−1)-forms) is a p p representative of a Dolbeault cohomology class ωϕ of bidegree (0,p−1) depending only on ϕ and f, i.e., we have a map O ∋ ϕ 7→ ωϕ = [α ,α ,...,α ] ∈ H(0,p−1)(Ω\Z) ⊕rp . 0 1 2 rp ∂¯ (cid:16) (cid:17) Note that these cohomology classes form a O -module and that ϕω1 = 0 ωϕ for ϕ ∈ O . 0 5 Let X be a subset of Cn and denote ω1 by ω. By Dp,q(X) we mean the space of all (p,q)-forms that have compact support on X. Definition 2.2. The residue Res : {ξ ∈ Dn,n−p(Ω);∂¯ξ = 0 close to Z} → C f 0 is given by ¯ (10) Res (ξ) = ∂ξ ∧ω. f Z The fact that Res is well-defined, i.e., does not depend on the choice f of representant of ω, is a direct consequence of Stokes’ theorem. We define multiplication with a holomorphic germ ϕ analogous to the case of the Grothendieck residue, i.e., ϕRes (ξ) = Res (ϕξ) and we thus f f get ϕRes (ξ) = Res (ϕξ) = ∂¯(ϕξ)∧ω = (∂¯(ξ))∧ϕω = ∂¯ξ ∧ωϕ. f f Z Z Z Remark 2.3. If Z consists of one single point we can rewrite Res in f a different way. Let ξ ∈ Dn,0 such that ∂¯ξ = 0 close to 0. Then there 0 ¯ exists a compact set D ⊂ Ω, with 0 in the interior, such that ∂ξ = 0 ˜ ˜ on D. If ξ is a holomorphic (n,0)-form that satisfy ξ = ξ on D we get ¯ ¯ ˜ ∂ξ ∧ω = ∂ξ ∧ω = − ξ ∧ω = − ξ ∧ω. Z ZCn\D Z∂D Z∂D We will use this in Example 2.6 below. The following theorem is the main result in this paper. Theorem 2.4. Assume that f ,...,f ∈ O and that the ideal J 1 m 0 generated by the f :s is Cohen-Macaulay. Then the following are equiv- i alent: (i) ϕ ∈ J (ii) ωϕ = 0 (iii) ϕRes = 0 f We postpone the proof to the next section. Remark 2.5. The operator ∇ was first introduced by Mats Anders- f son in [1] and was later used in several papers to define residue currents that coincide with the Coleff-Herrera product in the case of complete intersection. The advantage of using ∇ to define the residue Res is f f that much of the work in the proof of Theorem 2.4 is hidden in the construction of the cohomology classes ωϕ. Example 2.6. Consider the case when J = hf ,...,f i defines a com- 1 p plete intersection. It is well known, [3], that the Koszul complex with coefficients in O , i.e., the complex 0 0 → O ⊗ΛpE →δf ... →δf O ⊗Λ2E →δf O ⊗E →δf C → 0, 0 0 0 6 JOHANNESLUNDQVIST where E is acomplex vector space ofdimension pwith abasis e ,...,e 1 p and where δ is defined as f δ : O ⊗ΛkE → O ⊗Λk−1E, f 0 0 n δ (ψ ⊗e ∧...∧e ) = ψ (−1)j+1f ⊗e ∧...∧e ∧...∧e f l1 lk j l1 lj lk Xj=1 c is a minimal resolution of O /J. This means that the resolution (5) 0 is isomorphic to the Koszul complex since all minimal resolutions are isomorphic. In this case Lr(Ω \ Z) and ∇ in the total complex (9) f become Lr(Ω\Z) = E (Ω\Z,E ) and ∇ = δ −∂¯ 0,k+r k f f Mk where E is the trivial bundle Ω×ΛkE. We define the operator k ∩ : E (Ω\Z,E )×E (Ω\Z,E ) → E (Ω\Z,E ) 0,r k 0,s l 0,r+s k+l by letting dz ⊗e ∩dz ⊗e = dz ∧dz ⊗e ∧e . I J K L I K J L Let us try to calculate the cohomology class ω in this case. Let p ¯ f ⊗e σ = j=1 j j and v = σ∩(1+∂¯σ+(∂¯σ)∩2+...+(∂¯σ)∩(p−1)). P |f|2 Then v ∈ L−1(Ω\Z) and since ∇ σ = 1 and (∂¯σ)∩p = 0 we get that f ∇ v = 1. This means that a representative for the class ω is given by f v = σ∩(∂¯σ)∩(p−1), and by using that ( p f¯ ⊗e )∩2 = 0 we get that p j=1 j j P f¯ ⊗e ∩( ∂¯f¯ ⊗e )∩(p−1) j j j j v = . p P |Pf|2p ¯¯ ¯¯ ¯¯ ¯¯ Now, ∂f ⊗e ∩∂f ⊗e = ∂f ⊗ e ∩ ∂f ⊗ e for all j,k = 1,...,p j j k k k k j j ¯¯ ¯ and since ∂f = df we get k k (−1)j−1f¯df¯ ∧...∧df¯ ∧...∧df¯ ⊗e ∧...∧e j 1 j p 1 p v = p! . p P (|f |2 +...+|f |2)p 1 c n This shows that in the case of a complete intersection the residue coincide with the cohomological residue in [10] and together with Remark 2.3 this shows that Res indeed is a generalization of the Grothendieck f residue (3). 3. The proof of Theorem 2.4 ¯ Wewillneedaresultthatdescribeswhenwecansolvethe∂-equation in our situation and also a variant of Hartogs’ phenomenon. To prove those results we use an integral representation of smooth (p,q)-forms called Koppelman’s formula. 7 Let ∆ = {(z,z);z ∈ Cn} ⊂ Cn ×Cn and ∂|z|2 b(z) = . 2πi|z|2 A form s(ζ,z) in Ω × Ω on the form s(ζ,z) = s (ζ ,z )d(ζ − z ) j j j j j that satisfies 2πi sj(ζj,zj)(ζj −zj) = 1 outsidePthe diagonal ∆ and s(ζ,z) = b(ζ −z)Pin a neighborhood of ∆ is called an admissible form (in the sense of Andersson) [1]. For an admissable form s one can prove that K = s∧(∂¯s)n−1 is ∂¯-closed outside ∆. By K we mean the p,q component of K that has bidegree (p,q) in z and (n−p,n−q−1) in ζ. If f is a smooth (p,q)-form then for z ∈ D it has the representation ¯ ¯ f(z) = ∂ K (ζ,z)∧f(ζ)+ K (ζ,z)∧∂f(ζ)+ K (ζ,z)∧f(ζ). z p,q−1 p,q p,q Z Z Z ζ∈D ζ∈D ζ∈∂D This representation is referred to as Koppelman’s formula. If we want ¯ ¯ to solve the equation ∂u = f, where f is ∂-closed in some region D, Koppelman’s formula tells us that it is possible if we can make the boundary integral disappear. Remark 3.1. Koppelman’s formula is often stated so that the form s(ζ,z) is equal to b(ζ − z), see for example [6]. The formula above follows from the ordinary Koppelman’s formula. One way to see this is to first fix z ∈ D and then write f = χf + (1 − χ)f where χ is 0 a cutoff function with suppport in a small neighborhood U of z such 0 that s(ζ,z) = b(ζ−z) in U. The formula now follows fromthe ordinary ¯ Koppelman’s formula because of the ∂-closeness of K. Lemma 3.2. Write Cn = Cn−k × Ck and z = (z′,z′′), ζ = (ζ′,ζ′′). Assume that f is a ∂¯-closed smooth (0,q)-form in B = B′ ×B′′, where B′ and B′′ are the Euclidean (n−k) and k-balls, and that f has compact support in the z′′ direction. Then there exists a solution to ∂¯u = f in a possibly smaller set with compact support in the z′′ direction if q < k. If q = k such a solution exists if and only if (11) ξ ∧f = 0 Z for all ∂¯-closed (n,n−k)-forms ξ with compact support in the z′ direction. Proof. The ”only if” part of the statement when q = k is clear because if there is a solution u to ∂¯u = f with compact support in the z′′ direction then ¯ ¯ ξ ∧f = ξ ∧∂u = ∂(ξ ∧u) = 0 Z Z Z for all ∂¯-closed (n,n −k)-forms ξ with compact support in the z′ direction by Stokes’ theorem, since ξ ∧u has compact support. 8 JOHANNESLUNDQVIST Let χ′ be a cutoff function in B′ that is equal to 1 in a neighborhood of rB′, where r < 1 and let χ′′ be a cutoff function in B′′ that is equal to 1 in a neighborhood of rB′′. Set z¯′′ ·d(ζ −z) s(ζ,z) =χ′(ζ′) χ′′(z′′)b(ζ −z)+(1−χ′′(z′′)) + (cid:20) 2πi(|z′′|2 −ζ′′·z¯′′)(cid:21) ζ¯′ ·d(ζ −z) +(1−χ′(ζ′)) . (cid:20)2πi(|ζ′|2 −z′ ·ζ¯′)(cid:21) Then s(ζ,z) is admissible for |z′| ≤ r and for |ζ′′| ≤ r. Note that we can extend s to an admissible form for |ζ′′| < 1 simply by considering χs+(1−χ)b where χ is a cutoff function in rB. Since we can assume that χ is 1 in suppf this extension will be of no interest since K ∧f = 0 outside the suppport of f. This means that for our s, Koppelman’s formula will work for all z′′. If |ζ′| is close to 1 we get ζ¯′ ·d(ζ −z) s(ζ,z) = , 2πi(|ζ′|2 −z′ ·ζ¯′) which is holomorphic in z. Therefore the boundary integral in Koppel- man’s formula vanishes if q ≥ 0 since f has compact support in the ζ′′ ¯ direction. Thus u(z) = K ∧f is a solution to ∂u = f. It remains 0,q−1 to show that the solutioRn has compact support in the z′′ direction. Let |z′′| be close to 1. Then z¯′′ ·d(ζ −z) ζ¯′ ·d(ζ −z) s(ζ,z) = χ′(ζ′) +(1−χ′(ζ′)) (cid:20)2πi(|z′′|2 −ζ′′·z¯′′)(cid:21) (cid:20)2πi(|ζ′|2 −z′ ·ζ¯′)(cid:21) =: s (ζ,z)+s (ζ,z). 1 2 ¯ ¯ We see that ∂zs2 = 0 and that both s1 and s2 are ∂ζ′′-closed. This means that K = 0 if q < k because of degree reasons since then 0,q−1 n − q > n − k and K have bidegree (n,n − q) in ζ. In the case 0,q−1 ¯ q = k we will show that K is ∂ -closed and has compact support 0,q−1 ζ in the ζ′-direction. This will actually end the proof since then we can use (11) with ξ = K . 0,k−1 Assume q = k. Then K have bidegree (n,n − k) in ζ and thus 0,k−1 K is ∂¯ -closed since we get too many ζ′ differentials. Assume now 0,k−1 ζ that |ζ′| and |z′′| are close to 1. Then ζ¯′ ·d(ζ −z) s(ζ,z) = , 2πi(|ζ′|2 −z′ ·ζ¯′) and since it do not contain ζ′′,z′′,ζ¯′′ or ,z¯′′ we may regard it as an admissible form on B′ × B′. In particular, this means that K = s ∧ (∂¯s)n−k−1 is ∂¯-closed outside of ∆ which means that K = 0. (cid:3) 0,k−1 Proposition 3.3 (Variant of Hartogs’ phenomenon). Let Ω = Ω′×Ω′′, where Ω′′ hasdimensionk > 1, be an open setin Cn and letK = Ω′×rB 9 for some r < 1 such that rB ⊂ Ω′′. If q < k −1 then for each smooth ∂¯-closed (0,q)-form ν in (Ω \K) there exists a ∂¯-closed (0,q)-form νˆ in Ω such that νˆ = ν in Ω\Kˆ where Kˆ is a slightly bigger set than K. If q = 0 we have Kˆ = K. If q = k −1 the above statement is true if ¯ ∂ξ ∧ν = 0 Z for all (n,n−k)-forms ξ with compact support that are ∂¯-closed in a neighborhood of K. Proof. Let χ be a cutoff function in Ω that is identically 1 in a neigh- ¯ ¯ borhood of K and let g := (−∂χ)∧ν. Then g is ∂-closed in Ω and ¯ ¯ ξ ∧g = − ξ ∧∂χ∧ν = ± ∂(ξ ∧χ)∧ν = 0, Z Z Z for ∂¯-closed (n,n − q − 1)-forms ξ with compact support in the z′ direction. This means that we can use Lemma 3.2 with g as f and thus there exists a solution to ∂¯u = g, with compact support in the z′′ ¯ direction, in a possibly smaller set. Set νˆ = (1−χ)ν−u. Then ∂νˆ = 0 andνˆ = ν closetotheboundarywhere|z′′| = 1. Ifq = 0theuniqueness theorem for analytic functions imply that νˆ = ν in Ω\K. (cid:3) Proof. (Proof of Theorem 2.4) (i) ⇒ (ii): Assume that ϕ ∈ J. Let Ω be an open neighborhood of the origin such that L is exact for Ω\Z and such that there exist functions ψ ∈ O(Ω), such that j ϕ = ψ f . j j X Let {e } be a global frame of E such that f1(1 ⊗ e ) = f and let j 1 j j v = v + ... + v ∈ L−1 be defined by letting v = ψ ⊗ e and 1 p 1 j j v2 = v3 = ... = vp = 0. Then ∇fv1 = ϕ and ωϕ = 0 andPwe are done. (ii) ⇒ (iii): Trivial. (iii) ⇒ (i): Assume that ϕ ∈ O and that ϕRes = 0. Let again Ω be 0 f suchthatLisexactforΩ\Z andletv = v +v +...+v ∈ L−1(Ω\Z)be 1 2 p asolutionto∇v = ϕ. Becauseofgeneralpropertiesofcomplexanalytic sets we may assume that Ω is the set B′ × B′′, where B′ ⊂ Cn−p and B′′ ⊂ Cp aretheEuclidean balls, andthatZ donottouchtheboundary of rB′′ for some r < 1, [4]. According to Proposition 3.3 we can extend ¯ v to a ∂-closed form vˆ in Ω since v fulfills the requirement by the p p p ¯ assumption that ϕRes = 0. We can now solve the equation ∂u = vˆ f p p and since vˆ = v close to the boundary where |z′′| = 1 there exists a p p solutioninsayΩ\K where K isaset ofthesame typeasinProposition 3.3. Now, in Ω\K we get that ∂¯(v +fpu ) = fpv −fpvˆ = 0. p−1 p p p 10 JOHANNESLUNDQVIST This means that there exists a solution to ∂¯u = v +fpu in Ω\K p−1 p−1 p and we note that ∂¯(v +fp−1u ) = fp−1v +∂¯fp−1u = fp−1v −fp−1v p−2 p−1 p−1 p−1 p−1 p−1 = 0. If we repeat the argument above we eventually end up with ∂¯(v +f2u ) = 0 1 2 in a smaller set of the same type, call it U. Now, f1ψ = f1v +f1f2u = f1v = ϕ 1 2 1 in U and Proposition 3.3 in the case where q = 0 completes the proof. (cid:3) Acknowledgement: The author would like to thank professor Mats Andersson for suggesting the problem and for valuable comments on the early versions of the paper. The author would also like to thank the referee for his/her comments and suggestions which helped improve the paper. 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[9] Griffiths Ph., Harris J., Principles of Algebraic Geometry, Wiley, New York 1978. [10] Passare M., Residues, currents, and their relation to ideals of holomorphic functions, Math. Scand. 62 (1988), 75-152. Department of Mathematics Stockholm University SE-106 91 Stockholm Sweden E-mail address: [email protected]